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G 2. Problem: Let $\triangle A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$, and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c), $\left(c_{1}\right)$ be the circmucircles of the triangles $\triangle A E Z$ and $\triangle... |
Solution. Since the triangles $\triangle A E B$ and $\triangle C A B$ are similar, then
$$
\frac{A B}{E B}=\frac{C B}{A B}
$$
Since $A B=B Z$ we get
$$
\frac{B Z}{E B}=\frac{C B}{B Z}
$$
from which it follows that the triangles $\triangle Z B E$ and $\triangle C B Z$ are also similar. Since $F E B Z$ is cyclic,
!... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 115 |
G 3. Let $\triangle A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c) be the circumcircle of the triangle $\triangle A E Z$. Let $D$ be the second point of intersection of ... |
Solution. We will first show that $P A$ is tangent to $(c)$ at $A$.
Since $E, D, Z, A$ are concyclic, then $\angle E D C=\angle E A Z=\angle E A B$. Since also the triangles $\triangle A B C$ and $\triangle E B A$ are similar, then $\angle E A B=\angle B C A$, therefore $\angle E D C=\angle B C A$.
Since $\angle F E... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 116 |
NT 5. The positive integer $k$ and the set $A$ of different integers from 1 to $3 k$ inclusive are such that there are no distinct $a, b, c$ in $A$ satisfying $2 b=a+c$. The numbers from $A$ in the interval $[1, k]$ will be called small; those in $[k+1,2 k]$ - medium and those in $[2 k+1,3 k]$ - large. Is it always tr... |
Solution. A counterexample for a) is $k=3, A=\{1,2,9\}, x=2$ and $d=8$. A counterexample for c) is $k=3, A=\{1,8,9\}, x=8$ and $d=1$.
We will prove that b) is true.
Suppose the contrary and let $x, d$ have the above properties. We can assume $03 k$, then since the remainder for $x+d$ is medium we have $4 k2 k$. Ther... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 117 |
A 1. Let $x, y$ and $z$ be positive numbers. Prove that
$$
\frac{x}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}}}+\frac{y}{\sqrt{\sqrt[4]{z}+\sqrt[4]{x}}}+\frac{z}{\sqrt{\sqrt[4]{x}+\sqrt[4]{y}}} \geq \frac{\sqrt[4]{(\sqrt{x}+\sqrt{y}+\sqrt{z})^{7}}}{\sqrt{2 \sqrt{27}}}
$$
|
Solution. Replacing $x=a^{2}, y=b^{2}, z=c^{2}$, where $a, b, c$ are positive numbers, our inequality is equivalent to
$$
\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{\sqrt[4]{(a+b+c)^{7}}}{\sqrt{2 \sqrt{27}}}
$$
Using the Cauchy-Schw... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 119 |
A 3. Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{1}{a b(b+1)(c+1)}+\frac{1}{b c(c+1)(a+1)}+\frac{1}{c a(a+1)(b+1)} \geq \frac{3}{(1+a b c)^{2}}
$$
|
Solution. The required inequality is equivalent to
$$
\frac{c(a+1)+a(b+1)+b(c+1)}{a b c(a+1)(b+1)(c+1)} \geq \frac{3}{(1+a b c)^{2}}
$$
or equivalently to,
$$
(1+a b c)^{2}(a b+b c+c a+a+b+c) \geq 3 a b c(a b+b c+c a+a+b+c+a b c+1)
$$
Let $m=a+b+c, n=a b+b c+c a$ and $x^{3}=a b c$, then the above can be rewritten ... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 120 |
G 1. Let $H$ be the orthocentre of an acute triangle $A B C$ with $B C>A C$, inscribed in a circle $\Gamma$. The circle with centre $C$ and radius $C B$ intersects $\Gamma$ at the point $D$, which is on the arc $A B$ not containing $C$. The circle with centre $C$ and radius $C A$ intersects the segment $C D$ at the po... |
Solution. We use standard notation for the angles of triangle $A B C$. Let $P$ be the midpoint of $C H$ and $O$ the centre of $\Gamma$. As
$$
\alpha=\angle B A C=\angle B D C=\angle D K L
$$
the quadrilateral $A C K L$ is cyclic. From the relation $C B=C D$ we get $\angle B C D=180^{\circ}-2 \alpha$, so
$$
\angle A... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 123 |
G 2. Let $A B C$ be a right angled triangle with $\angle A=90^{\circ}$ and $A D$ its altitude. We draw parallel lines from $D$ to the vertical sides of the triangle and we call $E, Z$ their points of intersection with $A B$ and $A C$ respectively. The parallel line from $C$ to $E Z$ intersects the line $A B$ at the po... |
Solution. Suppose that the line $A A^{\prime}$ intersects the lines $E Z, B C$ and $C N$ at the points $L, M$, $F$ respectively. The line $I K$ being diagonal of the rectangle $K A^{\prime} I A$ passes through $L$, which by construction of $A^{\prime}$, is the middle of the other diagonal $A A^{\prime}$. The triangles... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 124 |
G 3. Let $A B C$ be an acute triangle, $A^{\prime}, B^{\prime}, C^{\prime}$ the reflexions of the vertices $A, B$ and $C$ with respect to $B C, C A$, and $A B$, respectively, and let the circumcircles of triangles $A B B^{\prime}$ and $A C C^{\prime}$ meet again at $A_{1}$. Points $B_{1}$ and $C_{1}$ are defined simil... |
Solution. Let $O_{1}, O_{2}$ and $O$ be the circumcenters of triangles $A B B^{\prime}, A C C^{\prime}$ and $A B C$ respectively. As $A B$ is the perpendicular bisector of the line segment $C C^{\prime}, O_{2}$ is the intersection of the perpendicular bisector of $A C$ with $A B$. Similarly, $O_{1}$ is the intersectio... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 125 |
G 4. Let $A B C$ be a triangle with side-lengths $a, b, c$, inscribed in a circle with radius $R$ and let $I$ be it's incenter. Let $P_{1}, P_{2}$ and $P_{3}$ be the areas of the triangles $A B I, B C I$ and $C A I$, respectively. Prove that
$$
\frac{R^{4}}{P_{1}^{2}}+\frac{R^{4}}{P_{2}^{2}}+\frac{R^{4}}{P_{3}^{2}} \... |
Solution. Let $r$ be the radius of the inscribed circle of the triangle $A B C$. We have that
$$
P_{1}=\frac{r c}{2}, \quad P_{2}=\frac{r a}{2}, \quad P_{3}=\frac{r b}{2}
$$
It follows that
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 126 |
G 5. Given a rectangle $A B C D$ such that $A B=b>2 a=B C$, let $E$ be the midpoint of $A D$. On a line parallel to $A B$ through point $E$, a point $G$ is chosen such that the area of $G C E$ is
$$
(G C E)=\frac{1}{2}\left(\frac{a^{3}}{b}+a b\right)
$$
Point $H$ is the foot of the perpendicular from $E$ to $G D$ an... |
Solution. Let $L$ be the foot of the perpendicular from $G$ to $E C$ and let $Q$ the point of intersection of the lines $E G$ and $B C$. Then,
$$
(G C E)=\frac{1}{2} E C \cdot G L=\frac{1}{2} \sqrt{a^{2}+b^{2}} \cdot G L
$$
So, $G L=\frac{a}{b} \sqrt{a^{2}+b^{2}}$.
=n^{2}+n+1$. Note that
$$
f\left(n^{2}\right)=n^{4}+n^{2}+1=\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)
$$
This means that $f(n) \mid f\left(n^{2}\right)$ for every positive integer $n$. By induction on $k$, one can easily see that $f(n) \mid f\left(n^{2^{k}}\right)$ for every positive integers $... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 129 |
A4 Real numbers $x, y, z$ satisfy
$$
0<x, y, z<1
$$
and
$$
x y z=(1-x)(1-y)(1-z) .
$$
Show that
$$
\frac{1}{4} \leq \max \{(1-x) y,(1-y) z,(1-z) x\}
$$
|
Solution: It is clear that $a(1-a) \leq \frac{1}{4}$ for any real numbers $a$ (equivalent to $0\max \{(1-x) y,(1-y) x,(1-z) x\}
$$
Now
$$
(1-x) y\frac{1}{2}$.
Using same reasoning we conclude:
$$
z\frac{1}{2}
$$
Using these facts we derive:
$$
\frac{1}{8}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}>x y z=(1-... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 130 |
A5 Let $x, y, z$ be positive real numbers. Prove that:
$$
\left(x^{2}+y+1\right)\left(x^{2}+z+1\right)\left(y^{2}+z+1\right)\left(y^{2}+x+1\right)\left(z^{2}+x+1\right)\left(z^{2}+y+1\right) \geq(x+y+z)^{6}
$$
|
Solution I: Applying Cauchy-Schwarz's inequality:
$$
\left(x^{2}+y+1\right)\left(z^{2}+y+1\right)=\left(x^{2}+y+1\right)\left(1+y+z^{2}\right) \geq(x+y+z)^{2}
$$
Using the same reasoning we deduce:
$$
\left(x^{2}+z+1\right)\left(y^{2}+z+1\right) \geq(x+y+z)^{2}
$$
and
$$
\left(y^{2}+x+1\right)\left(z^{2}+x+1\righ... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 131 |
G1 Let $A B C D$ be a parallelogram with $A C>B D$, and let $O$ be the point of intersection of $A C$ and $B D$. The circle with center at $O$ and radius $O A$ intersects the extensions of $A D$ and $A B$ at points $G$ and $L$, respectively. Let $Z$ be intersection point of lines $B D$ and $G L$. Prove that $\angle Z ... | ## Solution:
From the point $L$ we draw a parallel line to $B D$ that intersects lines $A C$ and $A G$ at points $N$ and $R$ respectively. Since $D O=O B$, we have that $N R=N L$, and point $N$ is the midpoint of segment $L R$.
Let $K$ be the midpoint of $G L$. Now, $N K \| R G$, and
$$
\angle A G L=\angle N K L=\an... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 132 |
G3 A parallelogram $A B C D$ with obtuse angle $\angle A B C$ is given. After rotating the triangle $A C D$ around the vertex $C$, we get a triangle $C D^{\prime} A^{\prime}$, such that points $B, C$ and $D^{\prime}$ are collinear. The extension of the median of triangle $C D^{\prime} A^{\prime}$ that passes through $... |
Solution: Let $A C \cap B D=\{X\}$ and $P D^{\prime} \cap C A^{\prime}=\{Y\}$. Because $A X=C X$ and $C Y=Y A^{\prime}$, we deduce:
$$
\triangle A B C \cong \triangle C D A \cong \triangle C D^{\prime} A^{\prime} \Rightarrow \triangle A B X \cong \triangle C D^{\prime} Y, \triangle B C X \cong \triangle D^{\prime} A^... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 133 |
G4 Let $A B C D E$ be a convex pentagon such that $A B+C D=B C+D E$ and let $k$ be a semicircle with center on side $A E$ that touches the sides $A B, B C, C D$ and $D E$ of the pentagon, respectively, at points $P, Q, R$ and $S$ (different from the vertices of the pentagon). Prove that $P S \| A E$.
|
Solution: Let $O$ be center of $k$. We deduce that $B P=B Q, C Q=C R, D R=D S$, since those are tangents to the circle $k$. Using the condition $A B+C D=B C+D E$, we derive:
$$
A P+B P+C R+D R=B Q+C Q+D S+E S
$$
From here we have $A P=E S$.
Thus,
$$
\triangle A P O \cong \triangle E S O\left(A P=E S, \angle A P O=... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 134 |
G5 Let $A, B, C$ and $O$ be four points in the plane, such that $\angle A B C>90^{\circ}$ and $O A=$ $O B=O C$. Define the point $D \in A B$ and the line $\ell$ such that $D \in \ell, A C \perp D C$ and $\ell \perp A O$. Line $\ell$ cuts $A C$ at $E$ and the circumcircle of $\triangle A B C$ at $F$. Prove that the cir... |
Solution: Let $\ell \cap A C=\{K\}$ and define $G$ to be the mirror image of the point $A$ with respect to $O$. Then $A G$ is a diameter of the circumcircle of the triangle $A B C$, therefore $A C \perp C G$. On the other hand we have $A C \perp D C$, and it implies that points $D, C, G$ are collinear.
Moreover, as $... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 135 |
C1 Inside of a square whose side length is 1 there are a few circles such that the sum of their circumferences is equal to 10 . Show that there exists a line that meets alt least four of these circles.
|
Solution
Find projections of all given circles on one of the sides of the square. The projection of each circle is a segment whose length is equal to the length of a diameter of this circle. Since the sum of the lengths of all circles' diameters is equal to $10 / \pi$, it follows that the sum of the lengths of all me... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 136 |
A1. Let $a, b, c, d, e$ be real numbers such that $a+b+c+d+e=0$. Let, also $A=a b+b c+c d+d e+e a$ and $B=a c+c e+e b+b d+d a$.
Show that
$$
2005 A+B \leq 0 \text { or } \quad A+2005 B \leq 0
$$
| ## Solution
We have
$$
0=(a+b+c+d+e)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+2 A+2 B
$$
This implies that
$$
A+B \leq 0 \text { or } 2006(\dot{A}+B)=(2005 A+B)+(A+2005 B) \leq 0
$$
This implies the conclusion.
| proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 137 |
G1. Let $A B C D$ be an isosceles trapezoid with $A B=A D=B C, A B / / D C, A B>D C$. Let $E$ be the point of intersection of the diagonals $A C$ and $B D$ and $N$ be the symmetric point of $\mathrm{B}$ with respect to the line $\mathrm{AC}$. Prove that quadrilateral $A N D E$ is cyclic.
| ## Solution
Let $\omega$ be a circle passing through the points $A, N, D$ and let $M$ the point where $\omega$ intersects $B D$ for the second time. The quadrilateral $A N D M$ is cyclic and it follows that
$$
\angle N D M+\angle N A M=\angle N D M+\angle B D C=180^{\circ}
$$
and
Figure 2
Assume that point $C$ lies on the line segment $B P$. By the Power of Point theorem we have $M A^{2}=M R \cdot M B$ and so $M P^{2}=M R \cdot M B$. The last equality i... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 139 |
G4. Let $\mathrm{ABC}$ be an isosceles triangle such that $A B=A C$ and $\angle \frac{A}{2}<\angle B$. On the extension of the altitude $\mathrm{AM}$ we get the points $\mathrm{D}$ and $\mathrm{Z}$ such that $\angle C B D=\angle A$ and $\angle Z B A=90^{\circ}$. $\mathrm{E}$ is the foot of the perpendicular from $\mat... | ## Solution
The points $A, B, K, Z$ and $C$ are co-cyclic.
Because ME//AC so we have
$$
\angle K E M=\angle E A C=\angle M B K
$$
Therefore the points $B, K, M$ and $E$ are co-cyclic. Now, we have
$$
\begin{aligned}
& \angle A B F=\angle A B C-\angle F B C \\
& =\angle A K C-\angle E K M=\angle M K C
\end{aligned}... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 140 |
G5. Let $A$ and $P$ are the points of intersection of the circles $k_{1}$ and $k_{2}$ with centers $O$ and $K$, respectively. Let also $B$ and $C$ be the symmetric points of $A$ with respect to $O$ and $K$, respectively. A line through $A$ intersects the circles $k_{1}$ and $k_{2}$ at the points $D$ and $E$, respectiv... | ## Solution
The points $B, P, C$ are collinear, and
$$
\angle A P C=\angle A P B=90^{\circ}
$$
Let $N$ be the midpoint of $D P$.
So we have:
$$
\begin{aligned}
& \angle N O P=\angle D A P \\
& =\angle E C P=\angle E C A+\angle A C P
\end{aligned}
$$
Since $O K / / B C$ and $O K$ is the bisector of $\angle A K P$ ... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 141 |
G7. Let $A B C D$ be a parallelogram, $\mathrm{P}$ a point on $C D$, and $Q$ a point on $A B$. Let also $M=A P \cap D Q, \quad N=B P \cap C Q, K=M N \cap A D$, and $L=M N \cap B C$. Show that $B L=D K$.
| ## Solution
Let $O$ be the intersection of the diagonals. Let $P_{1}$ be on $A B$ such that $P P_{1} / / A D$, and let $Q_{1}$ be on $C D$ such that $\mathrm{Q} Q_{1} / / A D$. Let $\sigma$ be the central symmetry with center $\mathrm{O}$. Let $\left.P^{\prime}=\sigma(P), Q^{\prime}=\sigma(Q), P_{1}^{\prime}=\sigma\le... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 142 |
NT3. Let $p$ be an odd prime. Prove that $p$ divides the integer
$$
\frac{2^{p!}-1}{2^{k}-1}
$$
for all integers $k=1,2, \ldots, p$.
| ## Solution
At first, note that $\frac{2^{p!}-1}{2^{k}-1}$ is indeed an integer.
We start with the case $\mathrm{k}=\mathrm{p}$. Since $p \mid 2^{p}-2$, then $p / 22^{p}-1$ and so it suffices to prove that $p \mid 2^{(p)!}-1$. This is obvious as $p \mid 2^{p-1}-1$ and $\left(2^{p-1}-1\right) \mid 2^{(p)!}-1$.
If $\m... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 143 |
NT5. Let $p$ be a prime number and let $a$ be an integer. Show that if $n^{2}-5$ is not divisible by $p$ for any integer $n$, there exist infinitely many integers $m$ so that $p$ divides $m^{5}+a$.
| ## Solution
We start with a simple fact:
Lemma: If $b$ is an integer not divisible by $p$ then there is an integer $s$ so that $s b$ has the remainder $l$ when divided by $p$.
For a proof, just note that numbers $b, 2 b, \ldots,(p-1) b$ have distinct non-zero remainders when divided by $p$, and hence one of them is ... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 144 |
C1. A triangle with area 2003 is divided into non-overlapping small triangles. The number of all the vertices of all those triangles is 2005 . Show that at mest one of the smaller triangles has area less or equal to 1.
| ## Solution
Since all the vertices are 2005 , and the vertices of the big triangle are among them, it follows that the number of the small triangles is at least 2003. So, it follows that at least one of the small triangles has area at most 1
| proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 145 |
A1 Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:
$\left(a^{5}+a^{4}+a^{3}+a^{2}+a+1\right)\left(b^{5}+b^{4}+b^{3}+b^{2}+b+1\right)\left(c^{5}+c^{4}+c^{3}+c^{2}+c+1\right) \geq 8\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)$.
| ## Solution
We have $x^{5}+x^{4}+x^{3}+x^{2}+x+1=\left(x^{3}+1\right)\left(x^{2}+x+1\right)$ for all $x \in \mathbb{R}_{+}$.
Take $S=\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)$.
The inequality becomes $S\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 8 S$.
It remains to pro... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 146 |
A2 Let $x, y, z$ be positive real numbers. Prove that:
$$
\frac{x+2 y}{z+2 x+3 y}+\frac{y+2 z}{x+2 y+3 z}+\frac{z+2 x}{y+2 z+3 x} \leq \frac{3}{2}
$$
| ## Solution 1
Notice that $\sum_{c y c} \frac{x+2 y}{z+2 x+3 y}=\sum_{c y c}\left(1-\frac{x+y+z}{z+2 x+3 y}\right)=3-(x+y+z) \sum_{c y c} \frac{1}{z+2 x+3 y}$.
We have to proof that $3-(x+y+z) \sum_{c y c} \frac{1}{z+2 x+3 y} \leq \frac{3}{2}$ or $\frac{3}{2(x+y+z)} \leq \sum_{c y c} \frac{1}{z+2 x+3 y}$.
By Cauchy-... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 147 |
A3 Let $a, b$ be positive real numbers. Prove that $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq a+b$.
|
Solution 1
Applying $x+y \leq \sqrt{2\left(x^{2}+y^{2}\right)}$ for $x=\sqrt{\frac{a^{2}+a b+b^{2}}{3}}$ and $y=\sqrt{a b}$, we will obtain $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq \sqrt{\frac{2 a^{2}+2 a b+2 b^{2}+6 a b}{3}} \leq \sqrt{\frac{3\left(a^{2}+b^{2}+2 a b\right)}{3}}=a+b$.
| proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 148 |
A9 Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying $\sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)$.
Prove that $\sum_{k=2}^{n-1} x_{k} \geq 0$.
| ## Solution 1
Case I. If $\min \left(x_{1}, x_{n}\right)=x_{1}$, we know that $x_{k} \geq \min \left(x_{k} ; x_{k+1}\right)$ for all $k \in\{1,2,3, \ldots, n-1\}$. So $x_{1}+x_{2}+\ldots+x_{n-1} \geq \sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)=x_{1}$, hence $\sum_{k=2}^{n-1} x_{k}... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 151 |
A7 Let $a, b$ and $c$ be a positive real numbers such that $a b c=1$. Prove the inequality
$$
\left(a b+b c+\frac{1}{c a}\right)\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right) \geq(1+2 a)(1+2 b)(1+2 c)
$$
| ## Solution 1
By Cauchy-Schwarz inequality and $a b c=1$ we get
$$
\begin{gathered}
\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(a b+b c+\frac{1}{c a}\right)}=\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(\frac{1}{c a}+a b+b c\right)} \geq \\
\left(\sqrt{a b} \cdot \sqrt{\frac{1}{a b}}+\sqrt{b c} \cdot \sqrt{b c}+\s... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 152 |
A8 Show that
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 4\left(\frac{x}{x y+1}+\frac{y}{y z+1}+\frac{z}{z x+1}\right)^{2}
$$
for any real positive numbers $x, y$ and $z$.
| ## Solution
The idea is to split the inequality in two, showing that
$$
\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}
$$
can be intercalated between the left-hand side and the right-hand side. Indeed, using the Cauchy-Schwarz inequality one has
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\fr... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 153 |
C1 On a $5 \times 5$ board, $n$ white markers are positioned, each marker in a distinct $1 \times 1$ square. A smart child got an assignment to recolor in black as many markers as possible, in the following manner: a white marker is taken from the board; it is colored in black, and then put back on the board on an emp... |
Solution
a) Position 20 white markers on the board such that the left-most column is empty. This
positioning is good because the coloring can be realized column by column, starting with the second (from left), then the third, and so on, so that the white marker on position $(i, j)$ after the coloring is put on positi... | proof | Combinatorics | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 154 |
C3 Integers $1,2, \ldots, 2 n$ are arbitrarily assigned to boxes labeled with numbers $1,2, \ldots, 2 n$. Now, we add the number assigned to the box to the number on the box label. Show that two such sums give the same remainder modulo $2 n$.
| ## Solution
Let us assume that all sums give different remainder modulo $2 n$, and let $S$ denote the value of their sum.
For our assumption,
$$
S \equiv 0+1+\ldots+2 n-1=\frac{(2 n-1) 2 n}{2}=(2 n-1) n \equiv n \quad(\bmod 2 n)
$$
But, if we sum, breaking all sums into its components, we derive
$$
S \equiv 2(1+\l... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 155 |
G1 Two perpendicular chords of a circle, $A M, B N$, which intersect at point $K$, define on the circle four arcs with pairwise different length, with $A B$ being the smallest of them.
We draw the chords $A D, B C$ with $A D \| B C$ and $C, D$ different from $N, M$. If $L$ is the point of intersection of $D N, M C$ a... | ## Solution
First we prove that $N L \perp M C$. The arguments depend slightly on the position of $D$. The other cases are similar.
From the cyclic quadrilaterals $A D C M$ and $D N B C$ we have:
$$
\varangle D C L=\varangle D A M \text { and } \varangle C D L=\varangle C B N \text {. }
$$
So we obtain
$$
\varangl... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 156 |
G2 For a fixed triangle $A B C$ we choose a point $M$ on the ray $C A$ (after $A$ ), a point $N$ on the ray $A B$ (after $B$ ) and a point $P$ on the ray $B C$ (after $C$ ) in a way such that $A M-B C=B N-A C=C P-A B$. Prove that the angles of triangle $M N P$ do not depend on the choice of $M, N, P$.
| ## Solution
Consider the points $M^{\prime}$ on the ray $B A$ (after $A$ ), $N^{\prime}$ on the ray $C B$ (after $B$ ) and $P^{\prime}$ on the ray $A C$ (after $C$ ), so that $A M=A M^{\prime}, B N=B N^{\prime}, C P=C P^{\prime}$. Since $A M-B C=B N-A C=B N^{\prime}-A C$, we get $C M=A C+A M=B C+B N^{\prime}=C N^{\pri... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 157 |
G4 Let $A B C$ be a triangle, $(B C<A B)$. The line $\ell$ passing trough the vertices $C$ and orthogonal to the angle bisector $B E$ of $\angle B$, meets $B E$ and the median $B D$ of the side $A C$ at points $F$ and $G$, respectively. Prove that segment $D F$ bisect the segment $E G$.
=a b \sin A$.
But it is also true that $(A B C D)=4(A O D)=4 \cdot \frac{O A \cdot O D}{2} \sin \theta=2 O A \cdot O D \sin \theta=$ $=2 \cdot \frac{x}{2}... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 160 |
G9 Let $O$ be a point inside the parallelogram $A B C D$ such that
$$
\angle A O B+\angle C O D=\angle B O C+\angle C O D
$$
Prove that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\triangle A O B, \triangle B O C, \triangle C O D$ and $\triangle D O A$.
 point on $\Gamma$. Denote by $\gamma$ the circle of diameter $A M$, by $X$ the (other than $M$ ) intersection... | ## Solution
Consider the line $\rho$ tangent to $\gamma$ at $A$, and take the points $\{K\}=A M \cap X Y,\{L\}=$ $\rho \cap X M$, and $\{F\}=O A \cap X Y$.
(Remark: Moving $M$ into its reflection with respect to the line $O A$ will move $X Y$ into its reflection with respect to $O A$. These old and the new $X Y$ meet... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 162 |
G11 Consider $A B C$ an acute-angled triangle with $A B \neq A C$. Denote by $M$ the midpoint of $B C$, by $D, E$ the feet of the altitudes from $B, C$ respectively and let $P$ be the intersection point of the lines $D E$ and $B C$. The perpendicular from $M$ to $A C$ meets the perpendicular from $C$ to $B C$ at point... |
Solution
Let $F$ be the foot of the altitude from $A$ and let $S$ be the intersection point of $A M$ and $R C$. As $P C$ is an altitude of the triangle $P R S$, the claim is equivalent to $R M \perp P S$, since the latter implies that $M$ is the orthocenter of $P R S$. Due to $R M \perp A C$, we need to prove that $A... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 163 |
NT2 Let $n \geq 2$ be a fixed positive integer. An integer will be called " $n$-free" if it is not a multiple of an $n$-th power of a prime. Let $M$ be an infinite set of rational numbers, such that the product of every $n$ elements of $M$ is an $n$-free integer. Prove that $M$ contains only integers.
| ## Solution
We first prove that $M$ can contain only a finite number of non-integers. Suppose that there are infinitely many of them: $\frac{p_{1}}{q_{1}}, \frac{p_{2}}{q_{2}}, \ldots, \frac{p_{k}}{q_{k}}, \ldots$, with $\left(p_{k}, q_{k}\right)=1$ and $q_{k}>1$ for each $k$. Let $\frac{p}{q}=\frac{p_{1} p_{2} \ldots... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 164 |
NT10 Prove that $2^{n}+3^{n}$ is not a perfect cube for any positive integer $n$.
| ## Solution
If $n=1$ then $2^{1}+3^{1}=5$ is not perfect cube.
Perfect cube gives residues $-1,0$ and 1 modulo 9 . If $2^{n}+3^{n}$ is a perfect cube, then $n$ must be divisible with 3 (congruence $2^{n}+3^{n}=x^{3}$ modulo 9 ).
If $n=3 k$ then $2^{3 k}+3^{2 k}>\left(3^{k}\right)^{3}$. Also, $\left(3^{k}+1\right)^{3... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 165 |
ALG 1. A number $A$ is written with $2 n$ digits, each of whish is 4 , and a number $B$ is written with $n$ digits, each of which is 8 . Prove that for each $n, A+2 B+4$ is a total square.
| ## Solution.
$$
\begin{aligned}
A & =\underbrace{44 \ldots 44}_{2 n}=\underbrace{44 \ldots 4}_{n} \underbrace{44 \ldots 4}_{n}=\underbrace{44 \ldots 4}_{n} \underbrace{400 \ldots 0}_{n}-\underbrace{44 \ldots 4}_{n}+\underbrace{88 \ldots 8}_{n}=\underbrace{44 \ldots 4}_{n} \cdot\left(10^{n}-1\right)+B \\
& =4 \cdot \un... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 166 |
ALG 2. Let $a, b, c$ be lengths of triangle sides, $p=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $q=\frac{a}{c}+\frac{c}{b}+\frac{b}{a}$.
Prove that $|p-q|<1$.
|
Solution: One has
$$
\begin{aligned}
a b c|p-q| & =a b c\left|\frac{c-b}{a}+\frac{a-c}{b}+\frac{b-a}{c}\right| \\
& =\left|b c^{2}-b^{2} c+a^{2} c-a c^{2}+a b^{2}-a^{2} b\right|= \\
& =\left|a b c-a c^{2}-a^{2} b+a^{2} c-b^{2} c+b c^{2}+a b^{2}-a b c\right|= \\
& =\left|(b-c)\left(a c-a^{2}-b c+a b\right)\right|= \\
... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 167 |
## ALG 4.
Let $a, b, c$ be rational numbers such that
$$
\frac{1}{a+b c}+\frac{1}{b+a c}=\frac{1}{a+b}
$$
Prove that $\sqrt{\frac{c-3}{c+1}}$ is also a rational number
|
Solution. By cancelling the denominators
$$
(a+b)^{2}(1+c)=a b+c\left(a^{2}+b^{2}\right)+a b c^{2}
$$
and
$$
a b(c-1)^{2}=(a+b)^{2}
$$
If $c=-1$, we obtrin the contradiction
$$
\frac{1}{a-b}+\frac{1}{b-a}=\frac{1}{a+b}
$$
Furtherrdore,
$$
\begin{aligned}
(c-3)(c+1) & =(c-1)^{2}-4=\frac{(a+b)^{2}}{a b}-4 \\
& =\... | proof | Algebra | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 169 |
ALG 6'. Let $a, b, c$ be positive numbers such that $a b+b c+c a=3$. Prove that
$$
a+b+c \geq a b c+2
$$
|
Solution. Eliminating $c$ gives
$$
a+b+c-a b c=a+b+(1-a b) c=a+b+\frac{(1-a b)(3-a b)}{a+b}
$$
Put $x=\sqrt{a b}$. Then $a+b \geq 2 x$, and since $1<x^{2}<3, \frac{(1-a b)(3-a b)}{a+b} \geq \frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x}$.
It then suffices to prove that
$$
2 x+\frac{\left(1-x^{2}\right)\left(... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 172 |
ALG 7 .
Let $x, y, z$ be real numbers greater than -1 . Prove that
$$
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2
$$
|
Solution. We have $y \leq \frac{1+y^{2}}{2}$, hence $\quad$
$$
\frac{1+x^{2}}{1+y+z^{2}} \geq \frac{1+x^{2}}{1+z^{2}+\frac{1+\dot{y}^{2}}{2}}
$$
and the similar inequalities.
Setting $a=1+x^{2}, b=1+y^{2}, c=1+z^{2}$, it sufices to prove that
$$
\frac{a}{2 c+b}+\frac{b}{2 a+c}+\frac{c}{2 b+a} \geq 1
$$
for all $a... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 173 |
ALG 8. Prove that there exist two sets $A=\{x, y, z\}$ and $B=\{m, n, p\}$ of positive integers greater than 2003 such that the sets have no common elements and the equalities $x+y+z=m+n+p$ and $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$ hold.
|
Solution. Let $A B C$ be a triangle with $B C=a, A C=b, A B=c$ and $ak+3=c
$$
a triangle with such length sides there exist. After the simple calculations we have
$$
\begin{gathered}
A=\left\{3(k+1)^{2}-2,3(k+2)^{2}+4,3(k+3)^{2}-2\right\} \\
B=\left\{3(k+1)^{2}, 3(k+2)^{2}, 3(k+3)^{2}\right\}
\end{gathered}
$$
It e... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 174 |
COM 3. Prove that amongst any 29 natural numbers there are 15 such that sum of them is divisible by 15 .
|
Solution: Amongst any 5 natural numbers there are 3 such that sum of them is divisible by 3 . Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisiblc by 3. It is easy to see ... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 175 |
COM 5. If $m$ is a number from the set $\{1,2,3,4\}$ and each point of the plane is painted in red or blue, prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$.
|
Solution. Suppose that in the plane there no exists an equilateral triangle with the vertices of the same colour and length side $m=1,2,3,4$.
First assertion: we shall prove that in the plane there no exists a segment with the length 2 such that the ends and the midpint of this segment have the same colour. Suppose t... | proof | Combinatorics | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 176 |
## GEO 3.
Let $G$ be the centroid of the the triangle $A B C$. Reflect point $A$ across $C$ at $A^{\prime}$. Prove that $G, B, C, A^{\prime}$ are on the same circle if and only if $G A$ is perpendicular to $G C$.
|
Solution. Observe first that $G A \perp G C$ if and only if $5 A C^{2}=A B^{2}+B C^{2}$. Indeed,
$$
G A \perp G C \Leftrightarrow \frac{4}{9} m_{a}^{2}+\frac{4}{9} m_{c}^{2}=b^{2} \Leftrightarrow 5 b^{2}=a^{2}+c^{2}
$$
Moreover,
$$
G B^{2}=\frac{4}{9} m_{b}^{2}=\frac{2 a^{2}+2 c^{2}-b^{2}}{9}=\frac{9 b^{2}}{9}=b^{2... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 177 |
GEO 5. Let three congruent circles intersect in one point $M$ and $A_{1}, A_{2}$ and $A_{3}$ be the other intersection points for those circles. Prove that $M$ is a.orthocenter for a triangle $A_{1} A_{2} A_{3}$.
|
Solution: The quadrilaterals $\mathrm{O}_{3} M O_{2} A_{1}, \mathrm{O}_{3} M O_{1} A_{2}$ and $O_{1} M O_{2} A_{3}$ are rombes. Therefore, $O_{2} A_{1} \| M O_{3}$ and $M O_{3} \| O_{1} A_{2}$, which imply $O_{2} A_{1} \| O_{1} A_{2}$. Because $O_{2} A_{1}=O_{3}{ }^{*} M=O_{1} A_{2}$ the quadrilateral $O_{2} A_{1} A_{... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 178 |
GEO 7. Through a interior point of a triangle, three lines parallel to the sides of the triangle are constructed. In that way the triangle is divided on six figures, areas equal $a, b, c, \alpha, \beta, \gamma$ (see the picture).
 Let $F$ and $G$ be the centers of the two circles of radius $R$ passing through $A$ and $B$; and $B$ and $C$, respectively. Let $O$ be the point for which the the rectangle $A B G O$ is a parallelogram. Then $\angle O A D=\angle G B C$, and the triangles $O A D$ and $G B C$ are congruent (sas... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 182 |
87.4. Let $a, b$, and $c$ be positive real numbers. Prove:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
|
Solution. The arithmetic-geometric inequality yields
$$
3=3 \sqrt[3]{\frac{a^{2}}{b^{2}} \cdot \frac{b^{2}}{c^{2}} \cdot \frac{c^{2}}{a^{2}}} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
or
$$
\sqrt{3} \leq \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}
$$
On the other ha... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 183 |
88.2. Let $a, b$, and $c$ be non-zero real numbers and let $a \geq b \geq c$. Prove the inequality
$$
\frac{a^{3}-c^{3}}{3} \geq a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$$
When does equality hold?
|
Solution. Since $c-b \leq 0 \leq a-b$, we have $(a-b)^{3} \geq(c-b)^{3}$, or
$$
a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \geq c^{3}-3 b c^{2}+3 b^{2} c-b^{3}
$$
On simplifying this, we immediately have
$$
\frac{1}{3}\left(a^{3}-c^{3}\right) \geq a^{2} b-a b^{2}+b^{2} c-b c^{2}=a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 184 |
90.1. Let $m, n$, and $p$ be odd positive integers. Prove that the number
$$
\sum_{k=1}^{(n-1)^{p}} k^{m}
$$
is divisible by $n$.
|
Solution. Since $n$ is odd, the sum has an even number of terms. So we can write it as
$$
\sum_{k=1}^{\frac{1}{2}(n-1)^{p}}\left(k^{m}+\left((n-1)^{p}-k+1\right)^{m}\right)
$$
Because $m$ is odd, each term in the sum has $k+(n-1)^{p}-k+1=(n-1)^{p}+1$ as a factor. As $p$ is odd, too, $(n-1)^{p}+1=(n-1)^{p}+1^{p}$ has... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 188 |
90.2. Let $a_{1}, a_{2}, \ldots, a_{n}$ be real numbers. Prove
$$
\sqrt[3]{a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}} \leq \sqrt{a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}}
$$
When does equality hold in (1)?
|
Solution. If $0 \leq x \leq 1$, then $x^{3 / 2} \leq x$, and equality holds if and only if $x=0$ or $x=1$. - The inequality is true as an equality, if all the $a_{k}$ 's are zeroes. Assume that at least one of the numbers $a_{k}$ is non-zero. Set
$$
x_{k}=\frac{a_{k}^{2}}{\sum_{j=1}^{n} a_{j}^{2}}
$$
Then $0 \leq x_... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 189 |
90.4. It is possible to perform three operations $f, g$, and $h$ for positive integers: $f(n)=$ $10 n, g(n)=10 n+4$, and $h(2 n)=n$; in other words, one may write 0 or 4 in the end of the number and one may divide an even number by 2. Prove: every positive integer can be constructed starting from 4 and performing a fi... |
Solution. All odd numbers $n$ are of the form $h(2 n)$. All we need is to show that every even number can be obtained fron 4 by using the operations $f, g$, and $h$. To this end, we show that a suitably chosen sequence of inverse operations $F=f^{-1}, G=g^{-1}$, and $H=h^{-1}$ produces a smaller even number or the num... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 190 |
91.4. Let $f(x)$ be a polynomial with integer coefficients. We assume that there exists a positive integer $k$ and $k$ consecutive integers $n, n+1, \ldots, n+k-1$ so that none of the numbers $f(n), f(n+1), \ldots, f(n+k-1)$ is divisible by $k$. Show that the zeroes of $f(x)$ are not integers.
|
Solution. Let $f(x)=a_{0} x^{d}+a_{1} x^{d-1}+\cdots+a_{d}$. Assume that $f$ has a zero $m$ which is an integer. Then $f(x)=(x-m) g(x)$, where $g$ is a polynomial. If $g(x)=b_{0} x^{d-1}+b_{1} x^{d-2}+$ $\cdots+b_{d-1}$, then $a_{0}=b_{0}$, and $a_{k}=b_{k}-m b_{k-1}, 1 \leq k \leq d-1$. So $b_{0}$ is an integer, and ... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 192 |
92.2. Let $n>1$ be an integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be $n$ different integers. Show that the polynomial
$$
f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1
$$
is not divisible by any polynomial with integer coefficients and of degree greater than zero but less than $n$ and s... |
Solution. Suppose $g(x)$ is a polynomial of degree $m$, where $1 \leq m<n$, with integer coefficients and leading coefficient 1 , such that
$$
f(x)=g(x) h(x)
$$
whre $h(x)$ is a polynomial. Let
$$
\begin{aligned}
& g(x)=x^{m}+b_{m-1} x^{m-1}+\cdots+b_{1} x+b_{0} \\
& h(x)=x^{n-m}+c_{n-m-1} x^{n-m-1}+\cdots+c_{1} x+... | proof | Algebra | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 193 |
94.1. Let $O$ be an interior point in the equilateral triangle $A B C$, of side length $a$. The lines $A O, B O$, and $C O$ intersect the sides of the triangle in the points $A_{1}, B_{1}$, and $C_{1}$. Show that
$$
\left|O A_{1}\right|+\left|O B_{1}\right|+\left|O C_{1}\right|<a
$$
|
Solution. Let $H_{A}, H_{B}$, and $H_{C}$ be the orthogonal projections of $O$ on $B C, C A$, and $A B$, respectively. Because $60^{\circ}\left|O A_{1}\right| \frac{\sqrt{3}}{2}
$$
In the same way,
$$
\left|O H_{B}\right|>\left|O B_{1}\right| \frac{\sqrt{3}}{2} \quad \text { and } \quad\left|O H_{C}\right|>\left|O C... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 195 |
94.3. A piece of paper is the square $A B C D$. We fold it by placing the vertex $D$ on the point $H$ of the side $B C$. We assume that $A D$ moves onto the segment $G H$ and that $H G$ intersects $A B$ at $E$. Prove that the perimeter of the triangle $E B H$ is one half of the perimeter of the square.
 The fold gives rise to an isosceles trapezium $A D H G$. Because of symmetry, the distance of the vertex $D$ from the side $G H$ equals the distance of the vertex
$H$ from side $A D$; the latter distance is the side length $a$ of the square. The line $G H$ thus is tangent to the circle with c... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 196 |
95.1. Let $A B$ be a diameter of a circle with centre $O$. We choose a point $C$ on the circumference of the circle such that $O C$ and $A B$ are perpendicular to each other. Let $P$ be an arbitrary point on the (smaller) arc $B C$ and let the lines $C P$ and $A B$ meet at $Q$. We choose $R$ on $A P$ so that $R Q$ and... |
Solution 1. (See Figure 7.) Draw $P B$. By the Theorem of Thales, $\angle R P B=\angle A P B=$ $90^{\circ}$. So $P$ and $Q$ both lie on the circle with diameter $R B$. Because $\angle A O C=90^{\circ}$, $\angle R P Q=\angle C P A=45^{\circ}$. Then $\angle R B Q=45^{\circ}$, too, and $R B Q$ is an isosceles right trian... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 197 |
95.3. Let $n \geq 2$ and let $x_{1}, x_{2}, \ldots x_{n}$ be real numbers satisfying $x_{1}+x_{2}+\ldots+x_{n} \geq 0$ and $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=1$. Let $M=\max \left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$. Show that
$$
M \geq \frac{1}{\sqrt{n(n-1)}}
$$
When does equality hold in (1)?
|
Solution. Denote by $I$ the set of indices $i$ for which $x_{i} \geq 0$, and by $J$ the set of indices $j$ for which $x_{j}<0$. Let us assume $M<\frac{1}{\sqrt{n(n-1)}}$. Then $I \neq\{1,2, \ldots, n\}$, since otherwise we would have $\left|x_{i}\right|=x_{i} \leq \frac{1}{\sqrt{n(n-1)}}$ for every $i$, and $\sum_{i=1... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 198 |
95.4. Show that there exist infinitely many mutually non-congruent triangles $T$, satisfying
(i) The side lengths of $T$ are consecutive integers.
(ii) The area of $T$ is an integer.
|
Solution. Let $n \geq 3$, and let $n-1, n, n+1$ be the side lengths of the triangle. The semiperimeter of the triangle then equals on $\frac{3 n}{2}$. By Heron's formula, the area of the triangle is
$$
\begin{gathered}
T=\sqrt{\frac{3 n}{2} \cdot\left(\frac{3 n}{2}-n+1\right)\left(\frac{3 n}{2}-n\right)\left(\frac{3 ... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 199 |
96.1. Show that there exists an integer divisible by 1996 such that the sum of the its decimal digits is 1996 .
|
Solution. The sum of the digits of 1996 is 25 and the sum of the digits of $2 \cdot 1996=3992$ is 23 . Because $1996=78 \cdot 25+46$, the number obtained by writing 781996 's and two 3992 in succession satisfies the condition of the problem. - As $3 \cdot 1996=5998$, the sum of the digits of 5988 is 30 , and $1996=65 ... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 200 |
96.3. The circle whose diameter is the altitude dropped from the vertex $A$ of the triangle $A B C$ intersects the sides $A B$ and $A C$ at $D$ and $E$, respectively $(A \neq D, A \neq E)$. Show that the circumcentre of $A B C$ lies on the altitude dropped from the vertex $A$ of the triangle $A D E$, or on its extensi... |
Solution. (See Figure 8.) Let $A F$ be the altitude of $A B C$. We may assume that $\angle A C B$ is sharp. From the right triangles $A C F$ and $A F E$ we obtain $\angle A F E=\angle A C F . \angle A D E$ and $\angle A F E$ subtend the same arc, so they are equal. Thus $\angle A C B=\angle A D E$, and the triangles $... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 201 |
97.2. Let $A B C D$ be a convex quadrilateral. We assume that there exists a point $P$ inside the quadrilateral such that the areas of the triangles $A B P, B C P, C D P$, and $D A P$ are equal. Show that at least one of the diagonals of the quadrilateral bisects the other diagonal.
 We first assume that $P$ does not lie on the diagonal $A C$ and the line $B P$ meets the diagonal $A C$ at $M$. Let $S$ and $T$ be the feet of the perpendiculars from $A$ and $C$ on the line $B P$. The triangles $A P B$ and $C B P$ have equal area. Thus $A S=C T$. If $S \neq T$, then the righ... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 203 |
97.4. Let $f$ be a function defined in the set $\{0,1,2, \ldots\}$ of non-negative integers, satisfying $f(2 x)=2 f(x), f(4 x+1)=4 f(x)+3$, and $f(4 x-1)=2 f(2 x-1)-1$. Show that $f$ is an injection, i.e. if $f(x)=f(y)$, then $x=y$.
|
Solution. If $x$ is even, then $f(x)$ is even, and if $x$ is odd, then $f(x)$ is odd. Moreover, if $x \equiv 1 \bmod 4$, then $f(x) \equiv 3 \bmod 4$, and if $x \equiv 3 \bmod 4$, then $f(x) \equiv 1 \bmod 4$. Clearly $f(0)=0, f(1)=3, f(2)=6$, and $f(3)=5$. So at least $f$ restricted to the set $\{0,1,2,3\}$ ia an inj... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 204 |
98.2. Let $C_{1}$ and $C_{2}$ be two circles intersecting at $A$ and $B$. Let $S$ and $T$ be the centres of $C_{1}$ and $C_{2}$, respectively. Let $P$ be a point on the segment $A B$ such that $|A P| \neq|B P|$ and $P \neq A, P \neq B$. We draw a line perpendicular to $S P$ through $P$ and denote by $C$ and $D$ the po... |
Solution. (See Figure 10.) The power of the point $P$ with respect to the circles $C_{1}$ and $C_{2}$ is $P A \cdot P B=P C \cdot P D=P E \cdot P F$. Since $S P$ is perpendicular to the chord $C D, P$
 \\
& \quad \geq\left(\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}}\right)\left(n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)
\end{aligned}
$$
When does e... |
Solution. The inequality of the problem can be written as
$$
\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}} \leq \frac{n\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)}{n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}
$$
A small manipulation of the right hand side brings the inequality to the equivalent form
$$
\frac{1}{\... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 207 |
00.3. In the triangle $A B C$, the bisector of angle $B$ meets $A C$ at $D$ and the bisector of angle $C$ meets $A B$ at $E$. The bisectors meet each other at $O$. Furthermore, $O D=O E$. Prove that either $A B C$ is isosceles or $\angle B A C=60^{\circ}$.
|
Solution. (See Figure 11.) Consider the triangles $A O E$ and $A O D$. They have two equal pairs of sides and the angles facing one of these pairs are equal. Then either $A O E$ and $A O D$ are congruent or $\angle A E O=180^{\circ}-\angle A D O$. In the first case, $\angle B E O=\angle C D O$, and
,(m+1, n),(m, n+1)$, and $(m+1, n+1)$ for some integers $m$ and $n$. Show that there exists a subcollection $B$ of $A$ such that $B$ contains at least $25 \%$ of the squares in $A... |
Solution. Divide the plane into two sets by painting the strips of squares parallel to the $y$ axis alternately red and green. Denote the sets of red and green squares by $R$ and $G$, respectively. Of the sets $A \cap R$ and $A \cap G$ at least one contains at least one half of the squares in $A$. Denote this set by $... | proof | Combinatorics | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 210 |
01.2. Let $f$ be a bounded real function defined for all real numbers and satisfying for all real numbers $x$ the condition
$$
f\left(x+\frac{1}{3}\right)+f\left(x+\frac{1}{2}\right)=f(x)+f\left(x+\frac{5}{6}\right)
$$
Show that $f$ is periodic. (A function $f$ is bounded, if there exists a number $L$ such that $|f(... |
Solution. Let $g(6 x)=f(x)$. Then $g$ is bounded, and
$$
\begin{gathered}
g(t+2)=f\left(\frac{t}{6}+\frac{1}{3}\right), \quad g(t+3)=f\left(\frac{t}{6}+\frac{1}{2}\right) \\
g(t+5)=f\left(\frac{t}{6}+\frac{5}{6}\right), \quad g(t+2)+g(t+3)=g(t)+g(t+5) \\
g(t+5)-g(t+3)=g(t+2)-g(t)
\end{gathered}
$$
for all real numbe... | proof | Algebra | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 211 |
01.4. Let $A B C D E F$ be a convex hexagon, in which each of the diagonals $A D, B E$, and $C F$ divides the hexagon in two quadrilaterals of equal area. Show that $A D, B E$, and $C F$ are concurrent.
 Denote the area of a figure by $|\cdot|$. Let $A D$ and $B E$ intersect at $P, A D$ and $C F$ at $Q$, and $B E$ and $C F$ at $R$. Assume that $P, Q$, and $R$ are different. We may assume that $P$ lies between $B$ and $R$, and $Q$ lies between $C$ and $R$. Both $|A B P|$ and $|D E P|$ differ ... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 212 |
02.3. Let $a_{1}, a_{2}, \ldots, a_{n}$, and $b_{1}, b_{2}, \ldots, b_{n}$ be real numbers, and let $a_{1}, a_{2}, \ldots, a_{n}$ be all different.. Show that if all the products
$$
\left(a_{i}+b_{1}\right)\left(a_{i}+b_{2}\right) \cdots\left(a_{i}+b_{n}\right)
$$
$i=1,2, \ldots, n$, are equal, then the products
$$... |
Solution. Let $P(x)=\left(x+b_{1}\right)\left(x+b_{2}\right) \cdots\left(x+b_{n}\right)$. Let $P\left(a_{1}\right)=P\left(a_{2}\right)=\ldots=P\left(a_{n}\right)=d$. Thus $a_{1}, a_{2}, \ldots, a_{n}$ are the roots of the $n$ :th degree polynomial equation $P(x)-d=0$. Then $P(x)-d=c\left(x-a_{1}\right)\left(x-a_{2}\ri... | proof | Algebra | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 214 |
03.1. Stones are placed on the squares of a chessboard having 10 rows and 14 columns. There is an odd number of stones on each row and each column. The squares are coloured black and white in the usual fashion. Show that the number of stones on black squares is even. Note that there can be more than one stone on a squ... |
Solution. Changing the order of rows or columns does not influence the number of stones on a row, on a column or on black squares. Thus we can order the rows and columns in such a way that the $5 \times 7$ rectangles in the upper left and lower right corner are black and the other two $5 \times 7$ rectangles are white... | proof | Combinatorics | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 215 |
03.3. The point $D$ inside the equilateral triangle $\triangle A B C$ satisfies $\angle A D C=150^{\circ}$. Prove that a triangle with side lengths $|A D|,|B D|,|C D|$ is necessarily a right-angled triangle.
 We rotate the figure counterclockwise $60^{\circ}$ around $C$. Because $A B C$ is an equilateral triangle, $\angle B A C=60^{\circ}$, so $A$ is mapped on $B$. Assume $D$ maps to $E$. The properties of rotation imply $A D=B E$ and $\angle B E C=150^{\circ}$. Because the triangle $D E C$ is eq... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 216 |
04.2. Let $f_{1}=0, f_{2}=1$, and $f_{n+2}=f_{n+1}+f_{n}$, for $n=1$, 2, ..., be the Fibonacci sequence. Show that there exists a strictly increasing infinite arithmetic sequence none of whose numbers belongs to the Fibonacci sequence. [A sequence is arithmetic, if the difference of any of its consecutive terms is a c... |
Solution. The Fibonacci sequence modulo any integer $n>1$ is periodic. (Pairs of residues are a finite set, so some pair appears twice in the sequence, and the sequence from the second appearance of the pair onwards is a copy of the sequence from the first pair onwards.) There are integers for which the Fibonacci resi... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 217 |
04.3. Let $x_{11}, x_{21}, \ldots, x_{n 1}, n>2$, be a sequence of integers. We assume that all of the numbers $x_{i 1}$ are not equal. Assuming that the numbers $x_{1 k}, x_{2 k}, \ldots, x_{n k}$ have been defined, we set
$$
\begin{aligned}
x_{i, k+1} & =\frac{1}{2}\left(x_{i k}+x_{i+1, k}\right), i=1,2, \ldots, n-... |
Solution. We compute the first index modulo $n$, i.e. $x_{1 k}=x_{n+1, k}$. Let $M_{k}=\max _{j} x_{j k}$ and $m_{k}=\min _{j} x_{j k}$. Evidently $\left(M_{k}\right)$ is a non-increasing and $\left(m_{k}\right)$ a non-decreasing sequence, and $M_{k+1}=M_{k}$ is possible only if $x_{j k}=x_{j+1, k}=M_{k}$ for some $j$... | proof | Algebra | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 218 |
04.4. Let $a, b$, and $c$ be the side lengths of a triangle and let $R$ be its circumradius. Show that
$$
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a} \geq \frac{1}{R^{2}}
$$
|
Solution 1. By the well-known (Euler) theorem, the inradius $r$ and circumradius $R$ of any triangle satisfy $2 r \leq R$. (In fact, $R(R-2 r)=d^{2}$, where $d$ is the distance between the incenter and circumcenter.) The area $S$ of a triangle can be written as
$$
A=\frac{r}{2}(a+b+c)
$$
and, by the sine theorem, as... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 219 |
05.2. Let $a, b$, and $c$ be positive real numbers. Prove that
$$
\frac{2 a^{2}}{b+c}+\frac{2 b^{2}}{c+a}+\frac{2 c^{2}}{a+b} \geq a+b+c
$$
|
Solution 1. Use brute force. Removing the denominators and brackets and combining simililar terms yields the equivalent inequality
$$
\begin{gathered}
0 \leq 2 a^{4}+2 b^{4}+2 c^{4}+a^{3} b+a^{3} c+a b^{3}+b^{3} c+a c^{3}+b c^{3} \\
-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}-2 a b c^{2}-2 a b^{2} c-2 a^{2} b c \\
=a^... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 220 |
05.4. The circle $\mathcal{C}_{1}$ is inside the circle $\mathcal{C}_{2}$, and the circles touch each other at $A$. A line through $A$ intersects $\mathcal{C}_{1}$ also at $B$ and $\mathcal{C}_{2}$ also at $C$. The tangent to $\mathcal{C}_{1}$ at $B$ intersects $\mathcal{C}_{2}$ at $D$ and $E$. The tangents of $\mathc... |
Solution. (See Figure 15.) Draw the tangent $\mathrm{CH}$ to $\mathcal{C}_{2}$ at $C$. By the theorem of the angle between a tangent and chord, the angles $A B H$ and $A C H$ both equal the angle at $A$ between $B A$ and the common tangent of the circles at $A$. But this means that the angles $A B H$ and $A C H$ are e... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 221 |
06.4. The squares of a $100 \times 100$ chessboard are painted with 100 different colours. Each square has only one colour and every colour is used exactly 100 times. Show that there exists a row or a column on the chessboard in which at least 10 colours are used.
|
Solution. Denote by $R_{i}$ the number of colours used to colour the squares of the $i$ 'th row and let $C_{j}$ be the number of colours used to colour the squares of the $j$ 'th column. Let $r_{k}$ be the number of rows on which colour $k$ appears and let $c_{k}$ be the number of columns on which colour $k$ appears. ... | proof | Combinatorics | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 222 |
07.2. A triangle, a line and three rectangles, with one side parallel to the given line, are given in such a way that the rectangles completely cover the sides of the triangle. Prove that the rectangles must completely cover the interior of the triangle.
|
Solution. Take any point $P$ inside the triangle and draw through $P$ the line parallel to the given line as well as the line perpendicular to it. These lines meet the sides of the triangle in four points. Of these four, two must be in one of the three rectangles. Now if the two points are on the same line, then the w... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 223 |
08.4. The difference between the cubes of two consecutive positive integers is a square $n^{2}$, where $n$ is a positive integer. Show that $n$ is the sum of two squares.
|
Solution. Assume that $(m+1)^{3}-m^{3}=n^{2}$. Rearranging, we get $3(2 m+1)^{2}=(2 n+$ $1)(2 n-1)$. Since $2 n+1$ and $2 n-1$ are relatively prime (if they had a common divisor, it would have divided the difference, which is 2 , but they are both odd), one of them is a square (of an odd integer, since it is odd) and ... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 226 |
09.4. There are 32 competitors in a tournament. No two of them are equal in playing strength, and in a one against one match the better one always wins. Show that the gold, silver, and bronze medal winners can be found in 39 matches.
|
Solution. To determine the gold medalist, we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners, 2 and finally one match, 31 matches altogether. Now the silver medal winner has at some point lost to number 1 ; as there were 5 rounds, there are 5 candidates. Let $C_{i}$ be the candid... | proof | Combinatorics | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 228 |
10.1. A function $f: \mathbb{Z} \rightarrow \mathbb{Z}_{+}$, where $\mathbb{Z}_{+}$is the set of positive integers, is non-decreasing and satisfies $f(m n)=f(m) f(n)$ for all relatively prime positive integers $m$ and $n$. Prove that $f(8) f(13) \geq(f(10))^{2}$.
|
Solution. Since $\mathrm{f}$ is non-decreasing, $f(91) \geq f(90)$, which (by factorization into relatively prime factors) implies $f(13) f(7) \geq f(9) f(10)$. Also $f(72) \geq f(70)$, and therefore $f(8) f(9) \geq f(7) f(10)$. Since all values of $\mathrm{f}$ are positive, we get $f(8) f(9) \cdot f(13) f(7) \geq$ $f... | proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 229 |
10.2. Three circles $\Gamma_{A}, \Gamma_{B}$ and $\Gamma_{C}$ share a common point of intersection $O$. The other common of $\Gamma_{A}$ and $\Gamma_{B}$ is $C$, that of $\Gamma_{A}$ and $\Gamma_{C}$ is $B$ and that of $\Gamma_{C}$ and $\Gamma_{B}$ is $A$. The
line $A O$ intersects the circle $\Gamma_{C}$ in the poin ... |
Solution 1. Let $\angle A O Y=\alpha, \angle A O Z=\beta$ and $\angle Z O B=\gamma$. So $\alpha+\beta+\gamma=180^{\circ}$. Also $\angle B O X=\alpha$ (vertical angles) and $\angle A C Y=\alpha=\angle B C X$ (angles subtending equal arcs); similarly $\angle C O X=\beta$, $\angle A B Z=\beta=\angle C B X ; \angle C O Y=... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 230 |
11.2. In a triangle $A B C$ assume $A B=A C$, and let $D$ and $E$ be points on the extension of segment $B A$ beyond $A$ and on the segment $B C$, respectively, such that the lines $C D$ and $A E$ are parallel. Prove that $C D \geq \frac{4 h}{B C} C E$, where $h$ is the height from $A$ in triangle ABC. When does equal... |
Solution. Because $A E \| D C$, the triangles $A B E$ and $D B C$ are similar. So
$$
C D=\frac{B C}{B E} \cdot A E
$$
$\mathrm{ja}$
$$
C D=\frac{A E \cdot B C}{B E \cdot C E} \cdot C E
$$
=x^{2}+x+1$. We have $P(n) P(n+1)=\left(n^{2}+n+1\right)\left(n^{2}+3 n+3\right)=$ $n^{4}+4 n^{3}+7 n^{2}+6 n+3$. Also, $P\left((n+1)^{2}\right)=n^{4}+4 n^{3}+7 n^{2}+6 n+3$. By choosing $a=(n+1)^{2}$ and $b=n+1$ we get $P(a) / P(b)=P(n)$ as desired.
| proof | Number Theory | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 234 |
Problem 2 Let $a, b, \alpha, \beta$ be real numbers such that $0 \leq a, b \leq 1$, and $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. Show that if
$$
a b \cos (\alpha-\beta) \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$
then
$$
a \cos \alpha+b \sin \beta \leq 1+a b \sin (\beta-\alpha)
$$
|
Solution 2 The condition can be rewritten as
$$
a b \cos (\alpha-\beta)=a b \cos \alpha \cos \beta+a b \sin \alpha \sin \beta \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$
Set $x=a \cos \alpha, y=b \sin \beta, z=b \cos \beta, t=a \sin \alpha$. We can now rewrite the condition as
$$
x z+y t \leq \sqrt{\lef... | proof | Inequalities | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 235 |
Problem 3 Let $M$ and $N$ be the midpoints of the sides $A C$ and $A B$, respectively, of an acute triangle $A B C, A B \neq A C$. Let $\omega_{B}$ be the circle centered at $M$ passing through $B$, and let $\omega_{C}$ be the circle centered at $N$ passing through $C$. Let the point $D$ be such that $A B C D$ is an i... |
Solution 3 Let $E$ be such that $A B E C$ is a parallelogram with $A B \| C E$ and $A C \| B E$, and let $\omega$ be the circumscribed circle of $\triangle A B C$ with centre $O$.
It is known that the radical axis of two circles is perpendicular to the line connecting the two centres. Since $B E \perp M O$ and $C E \... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 236 |
Problem 2. Given a triangle $A B C$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $B C$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $A B$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$... |
Solution I. There are three possibilities: $Q$ between $A$ and $B, Q=B$, and $B$ between $A$ and $Q$. If $Q=B$ we have that $\angle A B P$ is right, and $A P$ is a diameter
of the circumcircle. The triangles $A B P$ and $A C P$ are then congruent (they have $A P$ in common, $P B=P C$, and both have a right angle oppos... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 239 |
## Problem 1.
Let $A B C$ be a triangle and $\Gamma$ the circle with diameter $A B$. The bisectors of $\angle B A C$ and $\angle A B C$ intersect $\Gamma$ (also) at $D$ and $E$, respectively. The incircle of $A B C$ meets $B C$ and $A C$ at $F$ and $G$, respectively. Prove that $D, E, F$ and $G$ are collinear.
|
Solution 1. Let the line $E D$ meet $A C$ at $G^{\prime}$ and $B C$ at $F^{\prime} . A D$ and $B E$ intersect at $I$, the incenter of $A B C$. As angles subtending the same arc $\widehat{B D}$, $\angle D A B=\angle D E B=\angle G^{\prime} E I$. But $\angle D A B=\angle C A D=$ $\angle G^{\prime} A I$. This means that ... | proof | Geometry | proof | olympiads | false | nlile/NuminaMath-1.5-proofs-only | train | 240 |
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